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MULTIPLE OCCURRENCES OF BINOMIAL COEFFICIENTS CRAIG A . T O V E Y Georgia Institute of Technology, Atlanta, GA 30332 (Submitted February 1984) I. INTRODUCTION How many times can the same number appear in Pascalfs triangle? eliminating occurrences due to symmetry, (7 ) = ( (n\ occurrences of 1 = ( ln\ V ' Xn 7 " After ), and the uninteresting ; J and n = f J, the answer to this question is not clear. More precisely, if 1 < k < n/2, we say that ( -, 1 is a -proper binomial coefficient. Are there integers that can be expressed in different ways as proper binomial coefficients? Enumeration by hand or with a computer program produces some cases, given in Table 1. The smallest is 120, which equals (?) - (?)• Table 1. Small Multiple Occurrences of Binomial Coefficients INTEGER BINOMIAL COEFFICIENTS 120 Cs")- (?) 210 c:). (V) 1540 (?)• (?) 3003 i\% ('/)• (?) 7140 (?). en 11628 24310 (?)• (T) Cs7)- (T) There is even an instance of a number, 3003, which can be expressed in three different ways. No clear pattern emerges; the cases just seem to be sprinkled among the binomial coefficients. We conjecture that, for any t, there exist infinitely many integers that may be expressed in t different (proper) ways as binomial coefficients. 356 [Nov. MULTIPLE OCCURRENCES OF BINOMIAL COEFFICIENTS Here we prove the conjecture for the case t = 2. The proof is constructive and depends in an unexpected way on the Fibonacci sequence. I. THE CONSTRUCTION We seek solutions to an especially tractable situation because it leads to a second-order equation. In particular, if (1) holds, then n(k + 1 ) = (n - k)(n - k - 1). Let x = n - k - 1. (2) Then x(x + 1) = n{n - x) so n2 - xn - (x2 + x) = 0 and x + V5x2 + kx n= ,0>l (3^ ^ (since n is positive). lutions to (1). Integer solutions to (3) therefore lead to integer so- Since 5x2 + kx is even if and only if x is even, this means we must find integers x such that 5x2 + kx is a perfect square. Now x and 5x + 4 have no common factors except possibly 2 or 4, so a natural slightly stronger condition would be that both x and 5x + 4 be perfect squares. In other words, we need to find integers z such that 5z2 + 4 is a perfect square. These are given by the following lemma. Lemma 1: Let Fj denote the Fibonacci sequence. (F..x + F - + 1 ) 2 Then, for all J, - 5F| -4<-l)*. Proof: A straightforward calculation (see, e.g., [2], pp. 148-149) shows (.Fj + 1 + F..J* - 5F/ = 4 ^ + F.Fj_1 - F/) - -4(F? + F^^ - F?+1), which yields the result by induction. The lemma tells us that for any j even, z = Fj gives the perfect square +Fj+1)2. 5s 2 + 4 = {F._x This completes the construction. Theorem 1: Let Fj denote the Fibonacci sequence. Then, for any even j , there exists a solution to (1), where x = FJ and n and k are given by (2) and (3). Remark: Letting Lj denote the Lucas sequence as usual, we can write this solution as F jh +F I ,. F L J d -F! , Theorem 2: Theorem 1 gives all solutions to (1). Proof: It follows from the preceding discussion that any solution to (1) corresponds via (2) to some integer x such that 5x2+ kx is a perfect square. Let a (resp. b) be the number of times 2 divides 5x + 4 (resp. x). If a > 2, then 1985] 357 MULTIPLE OCCURRENCES OF BINOMIAL COEFFICIENTS b - 2 and, conversely, b > 2 implies a - 2. Since 5 ^ + 4 and x have no common factors except (possibly) 2 or 4, (5x + 4)/2 a is a perfect square, as is x/2b. Therefore, a + & is even, so a and 2? are both even or both odd. In the former case, x and 5x + 4 are perfect squares. We claim this leads precisely to the class of solutions given by Theorem 1. In the latter case, it follows that a = b = 1. Thus, we seek integers s such that 5s 2 + 2 is a perfect square. We further claim that no such integers exist. The two claims can be shown to follows from the general theory of the so-called Pell equation (see, for example, [1] for the first claim, and [3, pp. 350-358] for the second claim). For completeness, we give a simple proof that does not rely on the general theory. Let {An} denote any sequence of positive numbers satisfying the recurrence Tne An + An+1 = An+2' argument from Lemma 1 shows that, for all n, + (An-i A n+0 ~ ~>An - 4(An_1 + An_1An - An) + An = -\An + 2) + 5An + 1 . 2 2 Therefore, given any solution s, y to 5s + k - y , we can construct smaller solutions by setting y A A i S ~ > A ~ i-1 z " 2 i +1 ~ and extending the sequence {An} A n + A n+i = - A A 5 y + 2 z ' backward according to the recurrence A n+2* [The solutions will be z = Aj, y = AJ-_1 + Aj+li where j = i (mod 2), with \k\ = h\An + AnA2+1 - A2+1\, for all n.] Now, let (s, y) be any integer solution to 5z2 + 4 = z/2. ^ = 2 Then extend {An} If z > 3, then and ^- + 1 = — o — Set (an integer). backward to get the solution corresponding to A^_ ., ^ z/5 - z ^ . .61s ^ « ^ ^i-i = and A._ . y - % ^ zy/5 + .5 - z ^ _0 — 9 — 2 -72s, whence .28s < Ai_2 < .39s. Therefore, if Ai ^ 3, the solution corresponding to i4^_2 is smaller. Repeatedly extend {An} backward until Aj < 3. Since the only such integer solution is (1, 3 ) , s must have been a Fibonacci Number. This verifies the first claim. The second claim, that 5s 2 + 2 = y2 has no solutions, follows immediately from the fact that y2 = 2 (mod 5) has none. REFERENCES 1. 2. 3. 358 I.Gessel. Advanced Problem H-187. The Fibonacci Quarterly 10 (1972):417419. G. H. Hardy & E. M. Wright. An Introduction to the Theory of Numbers. Oxford: Oxford University Press, 1959. J. Uspensky & M. Heaslet. Elementary Number Theory. New York: McGraw-Hill, 1939. [Nov.